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Studentka2010 [4]
3 years ago
15

How do you know when a chemical equation is balanced?

Chemistry
1 answer:
kirza4 [7]3 years ago
6 0

Both sides of the equation would be equal.

Hope This Helps You!

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What properties differentiate an ionically bonded compound from a covalently bonded compound
zaharov [31]

Answer:

ionically bonded compound transfer its electrons while covalently bonded compound share their electrons. ionically bonded compounds are charged while covalently bonded compound are neutral

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3 years ago
A student uses red litmus paper to test the pH value of different solutions. Which would result in a color change of the litmus
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The answer is all of the above but fertilizer 
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The pH of a 0.25 M (aq) solution of hydrofluoric acid, HF, at 25C is 2.03. What is the value of Ka for HF? Pls show all mathemat
Snezhnost [94]

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Explanation:

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4 0
3 years ago
A covalent bond is best described as ______(A) a bond between two polyatomic ions.(B) a bond between a metal and a nonmetal.(C)
nekit [7.7K]

Answer:

Option C. the sharing of electrons between atoms

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Covalent bond is a type of bond in which the reacting element share their valence electrons in order to attain the noble gas configuration.

4 0
3 years ago
What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?
Ymorist [56]

Answer:

\% diss = 50\%

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}

Thus, it is possible to find x given the pH as shown below:

x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M

So that we can calculate the initial concentration of the acid:

\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\

[HA]_0=3.64x10^{-5}M

Therefore, the percent dissociation turns out to be:

\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%

Best regards!

6 0
2 years ago
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